已知y=y(x)由方程cos(x+y)+e^(x-y)=2 y' y'|(x,y)=(0,0)已知y=y(x)由方程cos(x+y)+e^(x-y)=2所确定,求y' ,y'|(x,y)=(0,0)
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已知y=y(x)由方程cos(x+y)+e^(x-y)=2 y' y'|(x,y)=(0,0)已知y=y(x)由方程cos(x+y)+e^(x-y)=2所确定,求y' ,y'|(x,y)=(0,0)
已知y=y(x)由方程cos(x+y)+e^(x-y)=2 y' y'|(x,y)=(0,0)
已知y=y(x)由方程cos(x+y)+e^(x-y)=2所确定,求y' ,y'|(x,y)=(0,0)
已知y=y(x)由方程cos(x+y)+e^(x-y)=2 y' y'|(x,y)=(0,0)已知y=y(x)由方程cos(x+y)+e^(x-y)=2所确定,求y' ,y'|(x,y)=(0,0)
对方程两边同时求导得,﹣﹙y+xy′﹚sin﹙xy﹚+e^y+﹙x+1﹚y′e^y=0
令x=0则方程cos(xy)+(x+1)*e^y=2为1+e^y=2,得y=0,即切点坐标为﹙0,0﹚
将﹙0,0﹚带入﹣﹙y+xy′﹚sin﹙xy﹚+e^y+﹙x+1﹚y′e^y=0得y′=﹣1
则过切点﹙0,0﹚的切线方程是y=﹣x,法线方程是y=x.
已知y=y(x)由方程cos(x+y)+e^(x-y)=2 y' y'|(x,y)=(0,0)已知y=y(x)由方程cos(x+y)+e^(x-y)=2所确定,求y' ,y'|(x,y)=(0,0)
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